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Mar 01, 2016
12:30 PM

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Mar 01, 2016
12:30 PM

Samme numbers, different results???

Can anyone please tell explain to me why these two calculations give two different results???

Solved! Go to Solution.

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Mar 01, 2016
12:58 PM

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Mar 01, 2016
12:58 PM

Search the forum for "temperature" and you'll get a lot of info.

Mathcad's base unit for temperature is Kelvin. Since 0°K does not equal 0°C, you end up with issues that need addressed.

If you clarify the meaning behind each of the temperatures in your equation, we'd be happy to help you sort it out. If you replace all the units with Δ°C, then you get the result you're expecting. However, that might not be the best way to represent whatever you're trying to calculate.

17 REPLIES 17

Mar 01, 2016
12:55 PM

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Mar 01, 2016
12:55 PM

300C=(300+273.15)K etc

Mar 01, 2016
12:58 PM

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Mar 01, 2016
12:58 PM

Could you elaborate on that? Can't see what Kelvin has to do with this!?

Mar 01, 2016
02:04 PM

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Mar 01, 2016
02:04 PM

Kelvin has everything to do with it. Most units convert with just a scaling factor, but temperature units do not. There's also an additive component. If you write 300C+1C what you usually mean is increase the temperature of 300C by 1C. That is, 300C+1deltaC, where 1deltaC=1K, as opposed to 1C=274.15K. There is no other way to handle temperature units in unit aware software, because the software cannot know what you mean by 300C+1C. Add 1K, or add 274.15K? The user has to distinguish between the two possibilities, because only the user knows what they want.

Mar 02, 2016
12:45 PM

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Mar 02, 2016
12:45 PM

Welcome to the wonderful world of empirical formulas.

Mar 02, 2016
01:14 PM

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Mar 02, 2016
01:14 PM

Fred Kohlhepp написал(а):

Welcome to the wonderful world of empirical formulas.

I see in reference books more pseudoempirical than empirical formulas!

Mar 01, 2016
12:58 PM

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Mar 01, 2016
12:58 PM

Search the forum for "temperature" and you'll get a lot of info.

Mathcad's base unit for temperature is Kelvin. Since 0°K does not equal 0°C, you end up with issues that need addressed.

If you clarify the meaning behind each of the temperatures in your equation, we'd be happy to help you sort it out. If you replace all the units with Δ°C, then you get the result you're expecting. However, that might not be the best way to represent whatever you're trying to calculate.

Mar 01, 2016
01:12 PM

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Mar 01, 2016
01:12 PM

It's a part of this formula where the fraction above is the one in the ln(...). So that specific fraction should just come out as a number without any unit. How can I do that without deleting the temperature units

Mar 01, 2016
01:13 PM

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Mar 01, 2016
01:13 PM

You're right, Δ°C does work and though it's not the best solution it might be the only one

Mar 01, 2016
01:48 PM

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Mar 01, 2016
01:48 PM

Without being familiar with this equation, it's hard to give much advice.

Firstly, are you certain that the temperatures should be in °C? You'll get a different result if you use another temperature unit (°K, °F, °R).

Secondly, what do the variables Θf, Θi, and β represent?

Mar 01, 2016
02:44 PM

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Mar 01, 2016
02:44 PM

β reciprocal conductor resistance temperature coefficient at 0 °C. β in °C.

Θi onset temperature in °C. Values can be taken from IEC 60287-3-1. If there is no defined values in the national tables should ambient temperatures of 20 °C at a depth of 1 m selected.

Θf final temperature in °C

Mar 02, 2016
10:56 AM

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Mar 02, 2016
10:56 AM

As expected, this is over my head.

If you don't like using Δ°C for all the units, you could consider omitting the units for these variables. For clarification, you could put °C in a text box beside the input variable.

Mar 02, 2016
12:05 PM

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Mar 02, 2016
12:05 PM

This could also be a solution. Thank you!

Mar 02, 2016
12:03 PM

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Mar 02, 2016
12:03 PM

Someone must be doing something weird with the units to get that equation, because the reciprocal of the resistance temperature coefficient would normally have units of °C/Ohm

Mar 01, 2016
01:02 PM

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Mar 01, 2016
01:02 PM

More interesting

Mar 01, 2016
01:09 PM

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Mar 01, 2016
01:09 PM

Mathcad 15

Mar 01, 2016
01:05 PM

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Mar 01, 2016
01:05 PM

You can't add temperatures in C (or F) like that.

300C=573.15K and 234.5C=507.65K. Therefore 300C+234.5C=1080.8K=807.65C.

When you add them mentally and get the result 534.5C what you are actually doing is removing the units to get just the numeric values, adding the numeric values, and then putting the units back on. Look up temperature units in the help.

Mar 01, 2016
01:12 PM

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